For a certain transverse standing wave on a long string, an antinode is atx-0 an

Question

For a certain transverse standing wave on a long string, an antinode is atx-0 and an adjacent node is at x-0.20 m. The displacement y(t) of the string particle at x-0 is shown in the figure, where the scale of the y axis is set by Ys 4.1 cm, when t 0.60 s, what is the displacement of the string particle at (a) x-0.40 m and (b) x-0.50 m ? What is the transverse velocity of the string particle at x-0.40 m at (c) t-0.60 s and (d) t 1.0 s? 0.5 Units m (a) Number-3.86 (b) Number 1-4.0 (c) Number 12.45 (d) Number8.745 Units m mbo Units m/s Units m/s

Explanation / Answer

distance between antinode and node = lambda / 4 = 0.20

lambda = 0.80 m

k = 2pi/k= 2.5 pi

w = 2 pi / T = pi

y = 4.1cm cos(k x - w t)

(A) t = 0.6 and x = 0.40 m

y = 1.3 cm

(B) t = 0.6 x = 0.50

y = - 1.86 cm

(C) v = dy/dt = 4.1pi cm/s sin(kx- wt)

v = 7.57 cm/s

(d) v = 0

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---